The equation of a circle $C$ is $x^2+y^2+12x-14y+49 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Explanation: To find the equation in standard form, complete the square. $(x^2+12x) + (y^2-14y) = -49$ $(x^2+12x+36) + (y^2-14y+49) = -49 + 36 + 49$ $(x+6)^{2} + (y-7)^{2} = 36 = 6^2$ Thus, $(h, k) = (-6, 7)$ and $r = 6$.